Stack & Quene

주식가격

1차 답안

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def solution(prices):
    answer = []
    for i in range(len(prices)):
        cnt = 0
        for j in range(i+1, len(prices)):
            if prices[i] <= prices[j]:
                cnt= cnt+1
        answer.append(cnt)
    return answer

비 효율성 문제로 통과 x

2차 답안

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def solution(prices):
    answer = [0] * len(prices)
    for i in range(len(prices)):
        for j in range(i+1, len(prices)):
            answer[i] += 1
            if prices[i] > prices[j]:
                break
    return answer

기능개발

1차 답안

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def solution(progresses, speeds):
    answer = []
    days = []
    for i in range(len(progresses)):
        days.append((100 - progresses[i]) / speeds[i])
    for j,k in enumerate(days):
        if j == 0:
            max = k
            answer.append(1)
            continue
        if k <= max:
            answer[-1] += 1
        else:
            max = k
            answer.append(1)
    return answer

케이스 2통과 불가..?

2차답안

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def solution(progresses, speeds):
    answer = []
    days = []
    
    for p,s in zip(progresses,speeds):
        if (100-p) % s > 0:
            days.append((100-p) // s + 1)
        else:
            days.append((100-p) // s)

    for j,k in enumerate(days):
        if j == 0:
            max = k
            answer.append(1)
            continue
        if k <= max:
            answer[-1] += 1
        else:
            max = k
            answer.append(1)
    return answer

소수점 관련 요소 수정 후 통과

다리를 지나는 트럭

1차 답안

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def solution(bridge_length, weight, truck_weights):  
    truck_in = []
    truck_out = []
    cnt = 0
    trucks_weight = 0
    n = len(truck_weights)

    while(len(truck_out) < n):
        idx = 0
        cnt = cnt + 1

        if len(truck_in) == 0:
            truck_in.append([1, truck_weights[0]])
            trucks_weight += truck_weights.pop(0)
        else:
            if len(truck_weights) == 0:
                for i in range(len(truck_in)):
                    truck_in[i][0] += 1

                    if truck_in[i][0] > bridge_length:
                        idx = 1

                    if idx == 1 and i == (len(truck_in) -1):
                        truck_out.append(truck_in.pop(0))

            else:
                for j in range(len(truck_in)):
                    truck_in[j][0] += 1

                    if truck_in[j][0] > bridge_length:
                        trucks_weight -= truck_in[j][1]
                        idx = 1

                    if idx == 1 and j == (len(truck_in) -1):
                        truck_out.append(truck_in.pop(0))

                    if len(truck_weights) == 0:
                        pass

                    elif (trucks_weight + truck_weights[0]) <= weight:
                        truck_in.append([1,truck_weights[0]])
                        trucks_weight += truck_weights.pop(0)
    return cnt

비효율 적인 코드여서 통과 불가

2차 답안

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def solution(bridge_length, weight, truck_weights):
    bridge = [0] * bridge_length
    sec = 0
    while bridge :
        bridge.pop(0)
        sec += 1
        if truck_weights :
            if (sum(bridge) + truck_weights[0]) <= weight :
                bridge.append(truck_weights.pop(0))
            else:
                bridge.append(0)
    return sec

sum(List)
List 요소의 합을 구하는 함수

프린터

1차 답안

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def solution(priorities, location):

    pi_list = [(p, i) for i, p in enumerate(priorities)]
    waiting_q = []

    while pi_list:
        pi = pi_list.pop(0)
        priority = pi[0]
        p_list = [priority for priority, idx in pi_list]
        if p_list:
            max_p = max(p_list)
        if priority >= max_p:
            waiting_q.append(pi)
        else:
            pi_list.append(pi)
    for i, item in enumerate(waiting_q):
        if item[1] == location:
            return i + 1